Shell Programming and Scripting

If you don't need to go down in the directory three and you don't want the relative path displayed, you can use Perl:

Code:

perl -le'print join ",",grep 1 > -M, <*>'

The above command will not display hidden files (add .* to the glob if you want to include them).

The reason you get the . is due to the find command.

The solution is not ideal as you will be getting the first . and an end ,

If you do the find command on its own you will see you get

.
./WS2003-WindowsMedia-BD.doc
./avcodec_sample.cpp

All the tr is doing is changing newline character \n to be ,

You may want to do something like

Code:

filelist=$(find . -type f -mtime 0 | tr '\n' ',')

If you are in ksh you could then strip off the last , of filelist with

Code:

filelist=${filelist%?}    #actually strips of last character

Thanks a lot for the timely help....

Hi all,

I am facing another issue with this find.....

Here is my program



#!/bin/sh
# Environmental set up
envSet="true";

# Initialize error code
error_code="false";

if [ "$envSet" = "$2" ]
then
# Set batch environment
. $1/jobs/setBatchEnv.sh $1
fi

echo "file name ---------"$filename
echo "BATCH_HOME ---------"$BATCH_HOME
echo "3 ---------"$3

#Frame the output file name
if [ "$filename" != "" ]
then
echo "Inside IF"
#if ZipReports.sh executed, then the filename of the zip will be taken from that
outfilename = $filename;
else
echo "Inside ELSE"
#if ZipReports.sh not executed, then the filename will be the list of files that are last modified today
cd $BATCH_HOME/data/output/$3
pwd
outfilename=$(find . -type f -mtime 0 | tr '\n' ',');
fi

echo "outifle-------"$outfilename

# Call the EmailReportsBatch.java program
$JAVA_HOME/java -Xmx1024m -Djava.awt.headless=true \
-DPROPERTIES_PATH=$BATCH_HOME/properties \
-DREPORT_TYPE=$3 \
-DOUTPUT_PATH=$BATCH_HOME/data/output/$3/ \
-DOUTPUT_FILE=$outfilename \
batch.EmailReportsBatch > $LOG_DIR/EmailReportsBatch.log

# Check if java command fails
if test ! $? -eq 0
then
error_code="true";
fi




When i execute this pgm i am getting the outfilename value correctly. But when it is invoked from main shell this outfilename is giving me one value less than what it is supposed to give. That is if 3 files are modified today, it is giving only 2 file names. Any suggestion why this is happening? But when BATCH_HOME, $3 values are echoed they show the same results...


Thanks,
Ananthi.U